IODS course project
Kisun Pokharel
Tools and Methods for open and reproducible research
The first week of the course covered basic tools and requirements for the rest of the weeks. I practiced basics of R by completing a Data Camp exercise called “R Short and Sweet”. In addition I brushed up git and rmarkdown by going through the first workshop as detailed below.
Rstudio Exercise 1
- I already had a github account and I just forked “IODS-project” from Tuomo Nieminen’s github.
- I modified chapter1.Rmd file under IODS-project in Rstudio. Link to my IODS-project repository is here.
- I opened the index.Rmd file and added my name. I commited the change and pushed to my github.
- I created a new Rmarkdown file with Rstudio. I wrote few sentences about the course and saved it as README.Rmd under IODS-project. The link to my course diary is here
- I replaced the default theme of my course diary web page with Time Machine theme.
- I uploaded all the changes to GitHub.
- I made sure that all changes could be seen.
Note1: Something that was not mentioned in the course that could be useful or could be included in future. Rstudio works hand-in-hand with github so R scripts or Rmd documents can be directly linked to github. The parent directory of the course (locally downloaded to the PC) can be imported as a github project that can be authorized by one’s username and password.
keywords: github, linux, rstudio, rmarkdown
Regression and Model validation
During the second week of this course, we have been delving deeper into R and statistics. We learned about regression models and the application of R in statistical modeling. The datacamp exercises along with the two embedded videos provided good background on the topics. Chapter three from “An Introduction to Statistical Learning with Applications in R” covered in-depth topics in linear regression.
RStudio Exercise 2
Data Wrangling
After going through the study materials, I attempted the RStudio exercise. The first part of the exercise was related to data wrangling where a subset of table was generated from a table with raw data (observations). The R script used to create the table can be found here.
Analysis
The R script for this part is available here. The data used in this exercise comes from an international survey of approaches to learning conducted by Kimmo Vehkalahti. The survey was funded by Teachers’ Academy funding (2013-2015) and the data was collected during December 2014 to January 2015. The survey was conducted in Finland with an aim to understand the relationship between learning approaches and students’ achievements in an introductory statistics course. A total of 183 individuals were included in the survey where the students were assessed for three different studying approaches - surface approach, deep approach and strategic approach. Additional details about the survey can be found here. After preprocessing in data wrangling step, we read the data into R and applied regression models.
lrn2014<-read.table("data/learning2014.csv")
str(lrn2014)## 'data.frame': 166 obs. of 7 variables:
## $ gender : Factor w/ 2 levels "F","M": 1 2 1 2 2 1 2 1 2 1 ...
## $ age : int 53 55 49 53 49 38 50 37 37 42 ...
## $ attitude: num 3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
## $ deep : num 3.58 2.92 3.5 3.5 3.67 ...
## $ stra : num 3.38 2.75 3.62 3.12 3.62 ...
## $ surf : num 2.58 3.17 2.25 2.25 2.83 ...
## $ points : int 25 12 24 10 22 21 21 31 24 26 ...dim(lrn2014)## [1] 166 7The final table used for analysis consist data including seven different variables and 166 individuals (see above). Among the variables, gender is a factor variable, age and points are integers whereas attitude, deep, stra and surf variables include float values.
summary(lrn2014)## gender age attitude deep stra
## F:110 Min. :17.00 Min. :1.400 Min. :1.583 Min. :1.250
## M: 56 1st Qu.:21.00 1st Qu.:2.600 1st Qu.:3.333 1st Qu.:2.625
## Median :22.00 Median :3.200 Median :3.667 Median :3.188
## Mean :25.51 Mean :3.143 Mean :3.680 Mean :3.121
## 3rd Qu.:27.00 3rd Qu.:3.700 3rd Qu.:4.083 3rd Qu.:3.625
## Max. :55.00 Max. :5.000 Max. :4.917 Max. :5.000
## surf points
## Min. :1.583 Min. : 7.00
## 1st Qu.:2.417 1st Qu.:19.00
## Median :2.833 Median :23.00
## Mean :2.787 Mean :22.72
## 3rd Qu.:3.167 3rd Qu.:27.75
## Max. :4.333 Max. :33.00The number of females (n=110) in this survey is almost two times the number of males (n=56). The age of students ranged from 17 years up to 55 years.
plot_lrn2014<-ggpairs(lrn2014, mapping = aes(col=gender, alpha = 0.3), lower=list(combo = wrap ("facethist", bins = 20)))
plot_lrn2014
The graphical overview of the data is shown above. Here, the overall goal of the survey is to identify how age of the students, attitude towards learning and three different learning methods are contributing towards the final points. In general, attitude towards learning has the highest impact for overall outcome of the study (i.e points scored) whereas deep learning method do not have any impact.
The explanatory variables were selected based on the absolute correlation values. The three explanatory variables for exam points (top correlated variables, also shown in the plot above) are student’s attitude towards learning (attitude), learning strategy (stra) and surface learning approach (surf). The model based on three dependent variables on exam points has the maximum residual value of 10.9 and median of 0.5. Here, residual value is the remaining value after the predicted value is substracted by the observed value. The model summary showed that attitude is highly significant (Pr=1.93e-08) variable that affects the student’s exam points. On the other hand, learning strategy and surface learning are not significant variables (Pr>0.01).
model<-lm(points ~ attitude + stra + surf, data = lrn2014)
summary(model)##
## Call:
## lm(formula = points ~ attitude + stra + surf, data = lrn2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.1550 -3.4346 0.5156 3.6401 10.8952
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.0171 3.6837 2.991 0.00322 **
## attitude 3.3952 0.5741 5.913 1.93e-08 ***
## stra 0.8531 0.5416 1.575 0.11716
## surf -0.5861 0.8014 -0.731 0.46563
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared: 0.2074, Adjusted R-squared: 0.1927
## F-statistic: 14.13 on 3 and 162 DF, p-value: 3.156e-08The summary of the model after removing insignificant variables is shown below. With regard to multiple r-squared value, we observed slight decrease in the value from 0.1927 (in earlier model) to 0.1856 (in updated model). However, other criteria for model evaluation such as F-Statistic (from 14.13 to 38.61) and p-value(3.156e-08 to 4.119e-09) have significantly improved. Thus, we can conclude that r-squared value alone may not determine the quality of the model. In this particular case, the lower r-squared value could be due to the outliers in the data.
model_sig<-lm(points ~ attitude, data = lrn2014)
summary(model_sig)##
## Call:
## lm(formula = points ~ attitude, data = lrn2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.9763 -3.2119 0.4339 4.1534 10.6645
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.6372 1.8303 6.358 1.95e-09 ***
## attitude 3.5255 0.5674 6.214 4.12e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared: 0.1906, Adjusted R-squared: 0.1856
## F-statistic: 38.61 on 1 and 164 DF, p-value: 4.119e-09par(mfrow = c(2,2))
plot(model_sig, which = c(1,2,5))
The three different diagnostic plots are generated above.The assumptions behind all three models is linearity and normality. Based on the above plots, we can conclude that the errors are normally distributed (clearly observed in q-q plot). Similarly, residual versus fitted model showed that the errors are not dependent on the attitude variable. Moreover, we can see that even two points (towards the right) have minor influence to the assumption in case of residual vs leverage model. All the models have adressed the outliers nicely. Thus, assumptions in all models are more or less valid.
Logistic Regression
One way to move on from linear regression is to consider settings where the dependent (target) variable is discrete. This opens a wide range of possibilities for modelling phenomena beyond the assumptions of continuity or normality.
Logistic regression is a powerful method that is well suited for predicting and classifying data by working with probabilities. It belongs to a large family of statistical models called Generalized Linear Models (GLM). An important special case that involves a binary target (taking only the values 0 or 1) is the most typical and popular form of logistic regression.
We will learn the concept of odds ratio (OR), which helps to understand and interpret the estimated coefficients of a logistic regression model. We also take a brief look at cross-validation, an important principle and technique for assessing the performance of a statistical model with another data set, for example by splitting the data into a training set and a testing set.
The slides and videos related to logistic regression can be found below.
Video: Logistic regression: probability and odds
Video: Logistic regression: Odds ratios
Video: Cross-validation: training and testing sets
Slides: Logistic regression
After going through the videos, we practiced DataCamp exercises on Logistic regression and started to work on the workshop (RStudio Exercise 3.
RStudio Exercise 3
Data Wrangling
The data for Exercise 3 was downloaded from UCI Machine Learning Repository (link). The zipped file contained two tables, namely student-mat.csv and student-por.csv. In this data wrangling exercise, the main task was to join two data sets and create a data frame for logistic regression analysis. More detailed information about the data is present in the next section (Data Analysis) of this exercise. The R script associated with this exercise can be found here
Data Analysis
The joined student alcohol consumption data that was created during wrangling exercise was read into R.
alc<-read.table("data/alc.csv")
#head(alc)
colnames(alc)## [1] "school" "sex" "age" "address" "famsize"
## [6] "Pstatus" "Medu" "Fedu" "Mjob" "Fjob"
## [11] "reason" "nursery" "internet" "guardian" "traveltime"
## [16] "studytime" "failures" "schoolsup" "famsup" "paid"
## [21] "activities" "higher" "romantic" "famrel" "freetime"
## [26] "goout" "Dalc" "Walc" "health" "absences"
## [31] "G1" "G2" "G3" "alc_use" "high_use"The data set in this exercise is a collection of information that is associated with student’s performance in two Portugese high schools. Two subjects - Mathematics (mat) and Portugese language (por) were choosen in this study. The findings from this study were published in the Proceedings of 5th FUture BUsiness TEChnology Conference (FUBUTEC 2008) during April, 2008 in Porto, Protugal(link). Altogether 33 attributes were assessed covering different aspects of student’s life. More detailed attribute information can be found here
Here, the main goal of the analysis is to study how alcohol consumption is associated with other aspects in student’s life. After going through the background information, it is a bit easier to identify interesting variables that could be related to alcohol consumption. Personally, I believe that the following are the four interesting variables that are associated with alcohol consumption:
Weekly study time (studytime) : In my opinion, if a student spends more time studying, he will have less time for alcohol consumption.
Going out with friends (goout) : In general, students go out with friends for parties and get-togethers. Attending such partiies and gatherings will lead higher alcohol consumption compared to those who do not participate in such activities.
Number of school absences (absences) : We can think of two reasons in terms of alcohol consumption and school absences. The main reason is that when a student consumes alcohol (especially during the evening), he/she will have lesser desire to go school next day (depends on the level of consumption). Another reason might be that a student is absent from class because he has plan to drink alcholic beverages.
Quality of family relationships (famrel) : I think the quality of family relationship will also affect student’s attitude towards alcohol consumption and the student who has bad family relationship may consume more alcohol compare to one who has better family relationship.
In the following section, we will see in details how my hypotheses are explained by the data. First, let’s summarise the subset of the table which includes the variables I have chosen.
library(dplyr)##
## Attaching package: 'dplyr'## The following object is masked from 'package:GGally':
##
## nasa## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionmy_var<- c("studytime", "absences", "goout", "famrel", "high_use")
my_var_data <- select(alc, one_of(my_var))
str(my_var_data)## 'data.frame': 382 obs. of 5 variables:
## $ studytime: int 2 2 2 3 2 2 2 2 2 2 ...
## $ absences : int 5 3 8 1 2 8 0 4 0 0 ...
## $ goout : int 4 3 2 2 2 2 4 4 2 1 ...
## $ famrel : int 4 5 4 3 4 5 4 4 4 5 ...
## $ high_use : logi FALSE FALSE TRUE FALSE FALSE FALSE ...All my chosen variables have integer values whereas information about the alcohol consumption is logical i.e True or False. Moreover, the str function also revealed the dimension of the selected data i.e 382 observations for five variables. After getting the data types, we can proceed with summarizing the table as follows:
summary(my_var_data)## studytime absences goout famrel
## Min. :1.000 Min. : 0.0 Min. :1.000 Min. :1.000
## 1st Qu.:1.000 1st Qu.: 1.0 1st Qu.:2.000 1st Qu.:4.000
## Median :2.000 Median : 3.0 Median :3.000 Median :4.000
## Mean :2.037 Mean : 4.5 Mean :3.113 Mean :3.937
## 3rd Qu.:2.000 3rd Qu.: 6.0 3rd Qu.:4.000 3rd Qu.:5.000
## Max. :4.000 Max. :45.0 Max. :5.000 Max. :5.000
## high_use
## Mode :logical
## FALSE:268
## TRUE :114
##
##
## The summary provides basic statistics (see the table above) about each variables. If we pick a particular variable absences (i.e number of absences), we can see that some of the students are never absent (min = 0)in the class whereas there have been a student or two who was absent upto 45 days (max = 0). Overall, when we look at the summary of all variables, median vlaues reflect better than mean values to understand the natures of our hypotheses i.e more vs less (studytime and absences, goout), high vs low, good vs bad (famrel). According to that, studying more than three hours, going out more than three times a week, being absent in class more than 3 days a week and having a relationship scale of more than 4 lead the students to upper levels and vice versa.
We can have graphical representation of each of the variables as bar charts (see below).
library(tidyr)
library(ggplot2)
gather(my_var_data) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar()
The summary tables provide information for alcohol consumption in relation to the factors for each selected variables.
t1 <- table("Study Time" = alc$studytime, "Alcohol Usage" = alc$high_use)
round(prop.table(t1, 1)*100, 1)## Alcohol Usage
## Study Time FALSE TRUE
## 1 58.0 42.0
## 2 69.2 30.8
## 3 86.7 13.3
## 4 85.2 14.8t2 <- table("Going Out" = alc$goout, "Alcohol Usage" = alc$high_use)
round(prop.table(t2, 1)*100, 1)## Alcohol Usage
## Going Out FALSE TRUE
## 1 86.4 13.6
## 2 84.0 16.0
## 3 81.7 18.3
## 4 50.6 49.4
## 5 39.6 60.4t3 <- table("Absences" = alc$absences, "Alcohol Usage" = alc$high_use)
round(prop.table(t3, 1)*100, 1)## Alcohol Usage
## Absences FALSE TRUE
## 0 80.0 20.0
## 1 74.5 25.5
## 2 72.4 27.6
## 3 80.5 19.5
## 4 66.7 33.3
## 5 72.7 27.3
## 6 76.2 23.8
## 7 75.0 25.0
## 8 70.0 30.0
## 9 50.0 50.0
## 10 71.4 28.6
## 11 33.3 66.7
## 12 50.0 50.0
## 13 50.0 50.0
## 14 14.3 85.7
## 16 0.0 100.0
## 17 0.0 100.0
## 18 50.0 50.0
## 19 0.0 100.0
## 20 100.0 0.0
## 21 50.0 50.0
## 26 0.0 100.0
## 27 0.0 100.0
## 29 0.0 100.0
## 44 0.0 100.0
## 45 100.0 0.0t4 <- table("Family Relationship" = alc$famrel, "Alcohol Usage" = alc$high_use)
round(prop.table(t4, 1)*100, 1)## Alcohol Usage
## Family Relationship FALSE TRUE
## 1 75.0 25.0
## 2 52.6 47.4
## 3 60.9 39.1
## 4 71.4 28.6
## 5 76.5 23.5Box plots provide more meaningful and summarized but more descriptive information for our variables as we can see the relationship of each four variables compared to alcohol consumption. Let’s look into more detail how the four variables I chose are affecting alcohol consumption in high school students using box plots.
library(ggpubr)## Loading required package: magrittr##
## Attaching package: 'magrittr'## The following object is masked from 'package:tidyr':
##
## extractg1 <- ggplot(alc, aes(x = high_use, y = studytime, col = high_use))
p1=g1 + geom_boxplot() + xlab("Alcohol Consumption")+ ylab("Study Time") + ggtitle("Study hours and alcohol consumption")
g2 <- ggplot(alc, aes(x = high_use, y = absences, col = high_use))
p2=g2 + geom_boxplot() + xlab("Alcohol Consumption")+ ylab("Number of School Absences") + ggtitle("School absences and alcohol consumption")
g3 <- ggplot(alc, aes(x = high_use, y = goout, col = high_use))
p3=g3 + geom_boxplot() + xlab("Alcohol Consumption")+ ylab("Going Out With Friends") + ggtitle("Going out with friends and alcohol consumption")
g4 <- ggplot(alc, aes(x = high_use, y = famrel, col = high_use))
p4=g4 + geom_boxplot() + xlab("Alcohol Consumption")+ ylab("Quality Family Relationship") + ggtitle("Family relationship and alcohol consumption")
ggarrange(p1, p2, p3 , p4, labels = c("A", "B", "C", "D"), ncol = 2, nrow = 2)
The four box plots above show how four of the chosen variables are associated with alcohol consumption. In each of the plots, x-axis shows the two different factors that measure the level of alcohol consumption i.e True for high consumption and False for low consumption and y-axis shows measurements for dependent variables i.e the four variables I have chosen. All of the box plots above show that what I hypothesized about the variables I selected in terms of alcohol consumption seem to be valid. We can see, they are valid but are these significantly valid observations? I will do a series of modeling and validations in the following sections.
Logistic Regression
Now we will do logistic regression where alcohol consumption (high_use) is target variable and four variables (studytime, goout, absences, famrel) I selected are the predictors.
m<-glm(high_use ~ studytime + goout + absences + famrel, data = alc, family = "binomial")
summary(m)##
## Call:
## glm(formula = high_use ~ studytime + goout + absences + famrel,
## family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.8701 -0.7738 -0.5019 0.8042 2.5416
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.28606 0.70957 -1.812 0.06992 .
## studytime -0.55089 0.16789 -3.281 0.00103 **
## goout 0.75953 0.12041 6.308 2.82e-10 ***
## absences 0.06753 0.02175 3.104 0.00191 **
## famrel -0.33699 0.13681 -2.463 0.01377 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 384.07 on 377 degrees of freedom
## AIC: 394.07
##
## Number of Fisher Scoring iterations: 4Among four variables, going out with friends (goout) is strongly coorelated (Pr = 2.82e-10) with alcohol consumption whereas quality of the family relationship (famrel) has comparatively lower impact towards alcohol consumption. Moreover, all four variables are significantly associated with alcohol consumption. Out of four variables, weekly study time (studytime) and the quality of family relationship (famrel) are inversely related to alcohol consumption. In other words, the more the number of hours spent in studying and the better the quality of the samily relationship, the lower the alcohol consumption. On the other hand, the number of school absences and frequency of going out with friends is positively correlated with alcohol consumption. This means that, if a student has higher number of school absences and goes out frequently with friends, his alcohol consumption is higher.
I will further delve into my model by evaluating it in terms of coefficients, odds ratio and confidence intervals.
Coef<-coef(m)
OR<-Coef %>% exp
CI<-confint(m) %>% exp## Waiting for profiling to be done...cbind(Coef, OR, CI)## Coef OR 2.5 % 97.5 %
## (Intercept) -1.28606058 0.2763573 0.06723596 1.0961732
## studytime -0.55089391 0.5764343 0.41040872 0.7941804
## goout 0.75953025 2.1372720 1.69853389 2.7261579
## absences 0.06753071 1.0698631 1.02591583 1.1187950
## famrel -0.33699130 0.7139151 0.54460646 0.9331198In general, If odds ratio is greater than 1, increase in explanotary variable will increase the response probability p whereas if odds ratio is less than 1, then increase in explanatory varialbe will decrease the response probability p. And if odds ratio equals to 1 then there is no effect of explanatory variable on response variable. According to these statements on odds ratios, the frequency of going out and the number of school absences have positive association with high alcohol usage. On the other hand, study time and family relationship seem to be negatively associated with high alcohol usage, because their odds ratio are smaller than one. With regards to confidence interval, the odds ratio for going out (goout) has the widest confidence interval i.e 2.73 (97.5%) and the study time has the narrowest confidence interval of 0.79 (97.5%). As none of the odds ratio have confidence interval of 1, we can claim that all of the explanatory variables have effect on the odds of outcome i.e. high alcohol usage.
Exploring predictive power of the model
To get insight into the predictive power of my model, I will compare model’s prediction with the probability of actual vaules.
#predict the probability
pred_prob <- predict(m, type = "response")
#add predicted probabilities and mutate the model
alc<-mutate(alc, probability = pred_prob)
#we will use the probability to validate the probabilities of choses four variables
alc<-mutate(alc, prediction = probability > 0.5)
table(high_use = alc$high_use, prediction = alc$prediction)## prediction
## high_use FALSE TRUE
## FALSE 242 26
## TRUE 65 49Based on the 2x2 cross-tabulation, we can see that my model predicted 65 false positives and 26 true negatives. In other words, prediction for a total of (65+26) 91 students is not true. To be more precise, we can check the overall percentage that the model is giving wrong prediction.
#first we need to define loss function
LF<-function(class, prob){
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
#now we compute the average number of wrong predictions
LF(alc$high_use, alc$prob)## [1] 0.2382199Now we can say that up to 24% of the predictions made by my model are false.
Cross Validation
Here we will perform 10-fold cross validation of our model
#load required library
library(boot)
CV<-cv.glm(data = alc, cost = LF, glmfit = m, K = 10 )
#finally look at the average number of wrong predictions
CV$delta[1]## [1] 0.2434555Now, after performing 10-fold cross validation, the perfomance of my model slightly increased. And yes, I can proudly claim that my model has better test set performance (24%) than the one we practised in data camp exercise(26%).
Clustering and Classification
The list of materials and links related to clustering and classification can be found below.
course slides by Emma Kämäräinen
DataCamp exercise
RStudio Exercise 4
After solving the DataCamp exercise and going through the embedded links, I got a general overview on the topic. In the following sections, I will prepare a report based on the exercise instructions. Unlike earlier weeks, the data wrangling exercise will be done after the data analysis part. In fact, the data wrangling exercise is part of Dimensionality Reduction Techniques. In the following section, I will explain about clustering and classification of data sets using open data called Boston that belongs to MASS package.
Data
First and foremost, it is important to get an overview of the data being analysed. As mentioned earlier, Boston data from MASS package.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## selectdata(Boston)
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...dim(Boston)## [1] 506 14The Boston data was collected to study the housing values in the suburbs of Boston. The table contains 506 observations for 14 different variables. The descriptions for each of the 14 variables are listed below.
| Variables | Description |
|---|---|
| crim | per capita crime rate by town. |
| zn | proportion of residential land zoned for lots over 25,000 sq.ft. |
| indus | proportion of non-retail business acres per town. |
| chas | Charles River dummy variable (= 1 if tract bounds river; 0 otherwise). |
| nox | nitrogen oxides concentration (parts per 10 million). |
| rm | average number of rooms per dwelling. |
| age | proportion of owner-occupied units built prior to 1940. |
| dis | weighted mean of distances to five Boston employment centres. |
| rad | index of accessibility to radial highways. |
| tax | full-value property-tax rate per $10,000. |
| ptratio | pupil-teacher ratio by town. |
| black | 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town. |
| lstat | lower status of the population (percent). |
| medv | median value of owner-occupied homes in $1000s. |
Data Summary
Now, let’s look at the summary of the boston data in the form of table (instead of default layout) using pandoc.table function of pander package.
library(pander)##
## Attaching package: 'pander'## The following object is masked from 'package:GGally':
##
## wrappandoc.table(summary(Boston), caption = "Summary of Boston data", split.table = 120)##
## -----------------------------------------------------------------------------------------------------------------------
## crim zn indus chas nox rm age
## ------------------ ---------------- --------------- ----------------- ---------------- --------------- ----------------
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000 Min. :0.3850 Min. :3.561 Min. : 2.90
##
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02
##
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000 Median :0.5380 Median :6.208 Median : 77.50
##
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917 Mean :0.5547 Mean :6.285 Mean : 68.57
##
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08
##
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000 Max. :0.8710 Max. :8.780 Max. :100.00
## -----------------------------------------------------------------------------------------------------------------------
##
## Table: Summary of Boston data (continued below)
##
##
## ------------------------------------------------------------------------------------------------------------------
## dis rad tax ptratio black lstat medv
## ---------------- ---------------- --------------- --------------- ---------------- --------------- ---------------
## Min. : 1.130 Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32 Min. : 1.73 Min. : 5.00
##
## 1st Qu.: 2.100 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38 1st Qu.: 6.95 1st Qu.:17.02
##
## Median : 3.207 Median : 5.000 Median :330.0 Median :19.05 Median :391.44 Median :11.36 Median :21.20
##
## Mean : 3.795 Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67 Mean :12.65 Mean :22.53
##
## 3rd Qu.: 5.188 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23 3rd Qu.:16.95 3rd Qu.:25.00
##
## Max. :12.127 Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90 Max. :37.97 Max. :50.00
## ------------------------------------------------------------------------------------------------------------------After getting a statistical summary of, it’s worthwhile to see to what extent each variables are correlated. For that, we use corr() function on Boston data.
library(corrplot)## corrplot 0.84 loadedlibrary(dplyr)
corr_boston<-cor(Boston) %>% round(2)
pandoc.table(corr_boston, split.table = 120)##
## -------------------------------------------------------------------------------------------------------------------------------
## crim zn indus chas nox rm age dis rad tax ptratio black lstat medv
## ------------- ------- ------- ------- ------- ------- ------- ------- ------- ------- ------- --------- ------- ------- -------
## **crim** 1 -0.2 0.41 -0.06 0.42 -0.22 0.35 -0.38 0.63 0.58 0.29 -0.39 0.46 -0.39
##
## **zn** -0.2 1 -0.53 -0.04 -0.52 0.31 -0.57 0.66 -0.31 -0.31 -0.39 0.18 -0.41 0.36
##
## **indus** 0.41 -0.53 1 0.06 0.76 -0.39 0.64 -0.71 0.6 0.72 0.38 -0.36 0.6 -0.48
##
## **chas** -0.06 -0.04 0.06 1 0.09 0.09 0.09 -0.1 -0.01 -0.04 -0.12 0.05 -0.05 0.18
##
## **nox** 0.42 -0.52 0.76 0.09 1 -0.3 0.73 -0.77 0.61 0.67 0.19 -0.38 0.59 -0.43
##
## **rm** -0.22 0.31 -0.39 0.09 -0.3 1 -0.24 0.21 -0.21 -0.29 -0.36 0.13 -0.61 0.7
##
## **age** 0.35 -0.57 0.64 0.09 0.73 -0.24 1 -0.75 0.46 0.51 0.26 -0.27 0.6 -0.38
##
## **dis** -0.38 0.66 -0.71 -0.1 -0.77 0.21 -0.75 1 -0.49 -0.53 -0.23 0.29 -0.5 0.25
##
## **rad** 0.63 -0.31 0.6 -0.01 0.61 -0.21 0.46 -0.49 1 0.91 0.46 -0.44 0.49 -0.38
##
## **tax** 0.58 -0.31 0.72 -0.04 0.67 -0.29 0.51 -0.53 0.91 1 0.46 -0.44 0.54 -0.47
##
## **ptratio** 0.29 -0.39 0.38 -0.12 0.19 -0.36 0.26 -0.23 0.46 0.46 1 -0.18 0.37 -0.51
##
## **black** -0.39 0.18 -0.36 0.05 -0.38 0.13 -0.27 0.29 -0.44 -0.44 -0.18 1 -0.37 0.33
##
## **lstat** 0.46 -0.41 0.6 -0.05 0.59 -0.61 0.6 -0.5 0.49 0.54 0.37 -0.37 1 -0.74
##
## **medv** -0.39 0.36 -0.48 0.18 -0.43 0.7 -0.38 0.25 -0.38 -0.47 -0.51 0.33 -0.74 1
## -------------------------------------------------------------------------------------------------------------------------------The table above shows the correlation matrix of all variables. Bird’s eye view on the matrix shows that tax (full-value property-tax rate) and rad (index of accessibility to radial highways) are the most positively correlated variables, whereas dis (weighted mean of distances to five Boston employment centres) and age (proportion of owner-occupied units built prior to 1940) are the most negatively correlated variables. Moreover, chas (Charles river dummy variable) and rad are the two variables that are least correlated.
The same information can be presented as a graphical overview. This time we will make a correlogram, a graphical representation of coorelation matrix. The corrplot() function of corrplot package wll be used to visualize the correlation between all the variables of the Boston dataset.
corrplot(corr_boston, method = "circle", tl.col = "black", cl.pos="b", tl.pos = "d", type = "upper" , tl.cex = 0.9 )
The above graph gives much quicker impression on which variables are more correlated to each other. In the graph, positive correlations are displayed in blue and negative correlations in red color with intensity of the color and circle size being proportional to the correlation coefficients. The same relationship as described above using correlation summary can be seen in the form of circles with different size (intensity of correlation i.e highly correlated or lowly correlated) and different colors (wheether positively or negatively correlated).
Data Standardization
Data scaling is useful for linear discriminant analysis. The scale() function will be used to scale the whole data. Here, the scaled value is generated by subtracting the column means from corresponding columns and then the difference is divided by standard deviation. i.e scaled(x)=(x-mean(x))/sd(x).
boston_scaled<-scale(Boston)
pandoc.table(summary(boston_scaled), caption = "Summary of Scaled Boston data", split.table = 120)##
## --------------------------------------------------------------------------------------------------------------
## crim zn indus chas nox rm
## ------------------- ------------------ ----------------- ----------------- ----------------- -----------------
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723 Min. :-1.4644 Min. :-3.8764
##
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681
##
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723 Median :-0.1441 Median :-0.1084
##
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
##
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823
##
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648 Max. : 2.7296 Max. : 3.5515
## --------------------------------------------------------------------------------------------------------------
##
## Table: Summary of Scaled Boston data (continued below)
##
##
## -----------------------------------------------------------------------------------------------------------
## age dis rad tax ptratio black
## ----------------- ----------------- ----------------- ----------------- ----------------- -----------------
## Min. :-2.3331 Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
##
## 1st Qu.:-0.8366 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
##
## Median : 0.3171 Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
##
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
##
## 3rd Qu.: 0.9059 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
##
## Max. : 1.1164 Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## -----------------------------------------------------------------------------------------------------------
##
## Table: Table continues below
##
##
## -----------------------------------
## lstat medv
## ----------------- -----------------
## Min. :-1.5296 Min. :-1.9063
##
## 1st Qu.:-0.7986 1st Qu.:-0.5989
##
## Median :-0.1811 Median :-0.1449
##
## Mean : 0.0000 Mean : 0.0000
##
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
##
## Max. : 3.5453 Max. : 2.9865
## -----------------------------------#corr_bostons<-cor(boston_scaled) %>% round(2)
#pandoc.table(corr_bostons, split.table = 120)We can make important observations on the summary of scaled data. The summary of the scaled Boston data has changed from the non-scaled Boston data. Most importantly, all the mean values have become zero and other values such as minimum, maximum, median and quartiles (1st and 3rd) are also changed for all variables.
Next, we will create quantile vector for crime using quantile function on scaled dataframe of Boston dataset. The quantile vectors will be labeled with meaningful labels to explain the intensity of crime i.e low, medium low, medium high and high. Lastly, we will replace the Crim variable with newly created crime variable and create the required data frame.
boston_scaled<- data.frame(boston_scaled)
qvc<-quantile(boston_scaled$crim)
crime <- cut(boston_scaled$crim, breaks = qvc, label = c("low", "med_low", "med_high", "high"), include.lowest = TRUE)
boston_scaled <- dplyr::select(boston_scaled, -crim)
boston_scaled<-data.frame(boston_scaled, crime)
#table(boston_scaled$crime)After creating the customized dataset in earlier steps, we will now divide it into training and testing sets where 80% of the data will belong to training set and 20% will be used as testing set.
#library(MASS)
n<-nrow(boston_scaled)
ind <- sample(n, size = n*0.8)
train <- boston_scaled[ind,]
test <- boston_scaled[-ind,]Now, as we have categorized the dataset into training and test set, we can fit linear discriminant analysis on the training set, where crime rate will be predicated based on all other variables.
Linear Discriminant Analysis
lda.fit <- lda(crime ~ ., data = train)
#add biplot arrows to an lda
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col=classes)
lda.arrows(lda.fit, myscale = 2)
# target classes as numeric
#classes <- as.numeric(train$crime)
# plot the lda results
#plot(lda.fit, dimen = 2, col = classes, pch = classes)Based on the bi-plot, it can be seen that rad variable alone acts as a predictor of high crime rate in the Boston data. On the other hand, the remaining 12 variables are associated with low, medium low and medium high rate of crime. The grouping based on 12 variables is fuzzy and is difficult to classify if any of the variables can classify the associated observations.
Class Prediction
crime_cat<-test$crime
test<-dplyr::select(test, -crime)
lda.pred<-predict(lda.fit, newdata = test)
table(correct = crime_cat, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 15 12 3 0
## med_low 3 13 5 0
## med_high 1 16 7 1
## high 0 0 0 26I tried to grasp the concept of the above matrix, which is also referred to as confusion matrix going through this blog. Everytime the matrix is generated, the number of correct and predicted cases for each of the classes (low, med_low, med_high, high) changes. The change is expected because of the randomized classification of test and training set. However, it was also observed that prediction for the high class fluctuated much lesser than the other classes.
K-means Clustering
In order to practice K-means clustering, we will reload the Boston data, scale the data and calculate the distances between the observations.
data(Boston)
boston_scaled1<-as.data.frame(scale(Boston))
dist_eu<-dist(boston_scaled1)
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970#head(boston_scaled1)We will use the scaled Boston data to perform K-means clustering. It’s always not trivial beforehand to identify how many clusters can classify our data. Therefore, we need to first randomly use certain number of clusters (if we can get any idea from the summary of the data or graphical summaries) but there are few other ways we can identify right number of clusters. This topics is more or less inspired by this R-blogger post and this Stackoverflow question
First we start with random cluster number. Let’s start with k=4 and apply k-means on the data.
#Let us apply kmeans for k=4 clusters
kmm = kmeans(boston_scaled1,6,nstart = 50 ,iter.max = 15) #we keep number of iter.max=15 to ensure the algorithm converges and nstart=50 to ensure that atleat 50 random sets are choosen Elbow method is also one of the well known techniques that can be used to estimate number of clusters.
#Elbow Method for finding the optimal number of clusters
library(ggplot2)
set.seed(1234)
# Compute and plot wss for k = 2 to k = 15.
k.max <- 15
data <- boston_scaled1
wss <- sapply(1:k.max,
function(k){kmeans(data, k)$tot.withinss})
#wss
qplot(1:k.max, wss, geom = c("point", "line"), span = 0.2,
xlab="Number of clusters K",
ylab="Total within-clusters sum of squares")## Warning: Ignoring unknown parameters: span
## Warning: Ignoring unknown parameters: span
Somehow the elbow plot shows that we may not see more than two clear clusters but it’s always nice to confirm such predictions using one more method because there is not shortage of methods for a number of analyses such as this. Therefore, we will additionally use NbClust package.
library(NbClust)
nb <- NbClust(boston_scaled1, diss=NULL, distance = "euclidean",
min.nc=2, max.nc=5, method = "kmeans",
index = "all", alphaBeale = 0.1)
## *** : The Hubert index is a graphical method of determining the number of clusters.
## In the plot of Hubert index, we seek a significant knee that corresponds to a
## significant increase of the value of the measure i.e the significant peak in Hubert
## index second differences plot.
## 
## *** : The D index is a graphical method of determining the number of clusters.
## In the plot of D index, we seek a significant knee (the significant peak in Dindex
## second differences plot) that corresponds to a significant increase of the value of
## the measure.
##
## *******************************************************************
## * Among all indices:
## * 12 proposed 2 as the best number of clusters
## * 6 proposed 3 as the best number of clusters
## * 3 proposed 4 as the best number of clusters
## * 3 proposed 5 as the best number of clusters
##
## ***** Conclusion *****
##
## * According to the majority rule, the best number of clusters is 2
##
##
## *******************************************************************#hist(nb$Best.nc[1,], breaks = max(na.omit(nb$Best.nc[1,])))Now, it’s much clearer that the data is described better with two clusters. With that, we run k-means algorithm again.
#Let us apply kmeans for k=4 clusters
km_final = kmeans(boston_scaled1, centers = 2) #we keep number of iter.max=15 to ensure the algorithm converges and nstart=50 to ensure that atleat 50 random sets are choosen
pairs(boston_scaled1[3:9], col=km_final$cluster)
The clusters in the above plot are divided into two groups and represented by two colors - red and black. Some of the pairs are better grouped than others in the plot. One of the important observations can be made with chas variable where the observations in all the pairs formed by it are wrongly clustered. On the other hand, clusters formed by rad variable are better separated.
More LDA
In the following section, we will use random cluster number (k=6) and perform LDA. We follow the the basic steps of scaling and distance calculation. Finally we will see how the biplot looks like on the whole data set when we try to group them into six categories.
boston_scaled2<-as.data.frame(scale(Boston))
#head(boston_scaled2)
set.seed(1234)
km_bs2<-kmeans(dist_eu, centers = 6)
#head(km_bs2)
myclust<-data.frame(km_bs2$cluster)
boston_scaled2$clust<-km_bs2$cluster
#head(boston_scaled2)
lda.fit_bs2<-lda(clust~., data = boston_scaled2 )
lda.fit_bs2## Call:
## lda(clust ~ ., data = boston_scaled2)
##
## Prior probabilities of groups:
## 1 2 3 4 5 6
## 0.10079051 0.19960474 0.09486166 0.20553360 0.12845850 0.27075099
##
## Group means:
## crim zn indus chas nox rm
## 1 -0.4149170 2.55535505 -1.228758914 -0.1951310 -1.21919439 0.78676843
## 2 0.3880377 -0.48724019 1.165421314 -0.2723291 0.98659851 -0.28553884
## 3 -0.3613809 -0.09419977 -0.474086929 1.5321752 -0.12487357 1.27068222
## 4 -0.3580718 -0.46023584 -0.003188584 -0.2723291 -0.09478548 -0.35414265
## 5 1.4172264 -0.48724019 1.069802298 0.4545202 1.34622349 -0.73713928
## 6 -0.4055840 0.02149547 -0.740804469 -0.2723291 -0.79649957 0.09099544
## age dis rad tax ptratio black
## 1 -1.4488239 1.7464736 -0.7048880 -0.5692695 -0.8353442 0.34924852
## 2 0.7651453 -0.7898745 1.1388129 1.2431405 0.6932747 0.04498348
## 3 0.2307707 -0.3386056 -0.4961654 -0.7220694 -1.1226766 0.32813467
## 4 0.4093998 -0.2612071 -0.5865335 -0.4342609 0.2608189 0.19191309
## 5 0.8557425 -0.9615698 1.2885597 1.2934457 0.4142248 -1.68787016
## 6 -0.8223904 0.7053125 -0.5694290 -0.7355910 -0.2013102 0.37698635
## lstat medv
## 1 -0.9773530 0.8760790
## 2 0.6734731 -0.5987824
## 3 -0.6138415 1.4407282
## 4 0.1508360 -0.2838601
## 5 1.1961180 -0.8078336
## 6 -0.5996059 0.2092896
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3 LD4 LD5
## crim 0.04811996 -0.28556378 -0.55488255 0.49400398 0.05329096
## zn -0.13738829 -1.83004313 0.34546140 -0.26802062 -0.87758918
## indus 0.74925386 -0.10015651 0.61607026 -0.42031079 0.25109137
## chas 0.13287282 -0.13228082 -0.94523359 -0.16829634 0.04786106
## nox 1.21764057 -0.81216848 -0.12506389 0.27633410 0.13213424
## rm -0.12060003 -0.04058521 -0.02502279 -0.75468374 0.21331834
## age 0.17397462 0.34382124 -0.07430813 -0.37956005 -0.95205471
## dis -0.36273454 -0.54652248 0.11546588 0.26210162 0.59195828
## rad 0.61453519 0.40958433 0.29006265 -0.40963042 1.56473994
## tax 0.75124298 -1.03741454 0.22707980 -0.17126395 -0.61781814
## ptratio 0.36217649 -0.18603253 0.30060517 0.16017164 -0.53729844
## black -0.27542772 0.27016025 0.77143821 -0.87012879 0.23445845
## lstat 0.48988940 -0.40861927 -0.53017288 -0.23295699 -0.06758426
## medv 0.22977036 -0.57759705 -0.86635437 -0.06977308 -0.10361245
##
## Proportion of trace:
## LD1 LD2 LD3 LD4 LD5
## 0.7285 0.1498 0.0750 0.0298 0.0168# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
plot(lda.fit_bs2, dimen = 2)
lda.arrows(lda.fit_bs2, myscale = 3)
I must admit that the number of clusters I chose was more than needed. I believe three to four clusters could group the whole data set. The top three most influential variables according to bi-plot are zn, nox and tax.
Better ways to visualize LDA
library(plotly)##
## Attaching package: 'plotly'## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:ggplot2':
##
## last_plot## The following object is masked from 'package:stats':
##
## filter## The following object is masked from 'package:graphics':
##
## layoutmodel_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crim)#Second 3D plot where colors are defined by clusters of k-means
#k-means_matpro<-kmeans(matrix_product, )
#head(train)
#train$cl<-myclust
#boston_scaled2$cl<-myclust
#head(boston_scaled2)
#head(train)
#rownames(train)
#rownames(boston_scaled2)
train$cl <- boston_scaled2$clust[match(rownames(train), rownames(boston_scaled2))]
#head(train)
#nrow(train)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type = "scatter3d", mode="markers", color = train$cl)Additional links (also included in the course slides)
Blog post by Jason Browniee on LDA
R-bloggers post on LDA R-bloggers post on K Means Clustering in R
Dimensionality Reduction Techniques
coming!